A button cell used in watches, functions as following $\begin{array}{r}\mathrm{Zn}\left(\mathrm{s}\right)+{\mathrm{Ag}}_2\mathrm{O}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons \\ 2\mathrm{Ag}\left(\mathrm{s}\right)+{\mathrm{Zn}}^{2+}(,)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)\end{array}$
If half-cell potentials are $\begin{array}{c}{\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}⟶\mathrm{Zn}\left(\mathrm{s}\right);{\mathrm{E}}^{\circ }=-0.76\mathrm{V}\\ {\mathrm{Ag}}_2\mathrm{O}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}⟶2\mathrm{Ag}\left(\mathrm{s}\right)+2{\mathrm{OH}}^{-}\end{array}$
${\mathrm{E}}^{\circ }=0.34\mathrm{V}$ the cell potential will be

Anode is always the site of oxidation and cathode is the site of reduction, Thus, anode half-cell is
${\mathrm{Zn}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}⟶\mathrm{Zn}\left(\mathrm{s}\right);{\mathrm{E}}^{\circ }=-0.76\mathrm{V}$
Cathode half-cell is ${\mathrm{Ag}}_2\mathrm{O}\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{e}}^{-}⟶2\mathrm{Ag}\left(\mathrm{s}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)$
$\begin{array}{l} {\mathrm{E}}^{\circ }=0.34\mathrm{V}\\ {\mathrm{E}}_{\text{cell }}^{\circ }={\mathrm{E}}_{\text{cathode }}^{\circ }-{\mathrm{E}}_{\text{anode }}^{\circ }\\ =0.34-(-0.76)\\ =1.1\mathrm{V}\end{array}$